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If Tan α = 1 1 + 2 − X and Tan β = 1 1 + 2 X + 1 Then Write the Value of α + β Lying in the Interval ( 0 , π 2 )

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Question

If tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\] 

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Solution

\[\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\]

\[ = \frac{\frac{1}{1 + 2^{- x}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{(1 + 2^{- x} )(1 + 2^{x + 1} )}}\]

\[ = \frac{1 + 2^{x + 1} + 1 + 2^{- x}}{1 + 2^{x + 1} + 2^{- x} + 2^{- x + x + 1} - 1}\]

\[ = \frac{2 + 2^{x + 1} + 2^{- x}}{2 + 2^{x + 1} + 2^{- x}}\]

\[ = 1\]

\[\text{ Therefore }, \alpha + \beta = \tan^{- 1} (1) = \frac{\pi}{4} .\]

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Chapter 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.3 [Page 27]

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R.D. Sharma Mathematics [English] Class 11
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.3 | Q 12 | Page 27

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